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Let's say I have following three sinusoidal waves.

 1. x1(t) = A*sin(2*pi*1000*t) i.e, f = 1 kHz   
 2. x2(t) = B*sin(2*pi*1500*t) i.e, f = 1.5 kHz   
 3. x3(t) = C*sin(2*pi*2000*t) i.e, f = 2 kHz

So when I get the signal y(t) = x1(t) + x2(t) + x3(t),

Is there any way to determine the common frequency such that the y(t) will start repeating itself? Can it be directly evaluated as smallest common multiple of these frequencies? For example, 6 kHz for this case.

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    \$\begingroup\$ Try plotting them. \$\endgroup\$
    – Chris Stratton
    Dec 12, 2019 at 18:00

2 Answers 2

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The interval of repetition will be the LCM (least common multiple) of the periods of the sinusoids. This is also the inverse of the GCD (greatest common divisor) of the frequencies.

The periods of your signals are 1ms, 0.666ms, and 0.5ms. The LCM of these is 2ms, so the signal repeats after 2 periods of the first signal, 3 periods of the second signal, and 4 periods of the third signal.

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  • \$\begingroup\$ Talking about LCM is useful in frequency, but less so in period times. After all, smallest frequency have highest period times. In period times, it would be the HCM (highest common multiple). \$\endgroup\$
    – Mast
    Dec 13, 2019 at 12:00
  • \$\begingroup\$ This "answer" actually doesn't answer the question "Is there something as a common frequency? ". As Bhuvnesh stated there is only if the frequency ratios are rational numbers (theoretically/mathematically speaking); only if the frequency sources are coupled (e. g. by PLL or by deriving from a common clock source; practically speaking) \$\endgroup\$
    – Curd
    Dec 13, 2019 at 13:12
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    \$\begingroup\$ @Curd I would argue that it is sufficient to demonstrate the existence of something by providing an extant example of that thing. If the frequency ratios are not rational then there will be no LCM or GCD, so I think the "only if" is strongly implied. \$\endgroup\$
    – Elliot Alderson
    Dec 13, 2019 at 13:42
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    \$\begingroup\$ Of course it is implied; but it is not evident. I consider it quite important to explicitely mention that in many cases something called "common frequency" doesn't exist or doesn't make sense. Especially since that was the question and OP picked one very special example that oversimplifies reality. \$\endgroup\$
    – Curd
    Dec 13, 2019 at 14:11
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It turns out that for such combinations of periodic signals:

Resulting Frequency will be equal to the Highest Common Factor of the all the frequencies .

Resulting Time Period will be equal to the Least Common Multiple of the all the Time Periods .

Highest common factor for 1000Hz, 1500Hz and 2000Hz is 500Hz. Hence Time Periodic equals 1/500= 2ms. Or you could go the other way round, calculating the Time period first using the Least common Multiple. Go for the easier option.

But this isn't always true. There may be signals whose combination can never give a periodic function. For example consider X1(t) with frequency, f1= π Hz and X2(t) with frequency, f2 =1 Hz. There is no finite number of cycles after which the combination of the two signals repeats, i.e their combination can't be periodic. It so turns out that for a combination of signals to be able to give a periodic function, the ratio of time periods(or frequencies) of the constituent functions must be rational. In my example it isn't rational, so resultant signal cannot be periodic. In your example the ratios are all rational and hence the resultant signal, y(t) is periodic.

You can satisfy yourself with these results by plotting the signals.

enter image description here

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